{"id":31503,"date":"2026-05-13T09:48:50","date_gmt":"2026-05-13T04:18:50","guid":{"rendered":"https:\/\/mycbseguide.com\/blog\/?p=31503"},"modified":"2026-05-13T11:50:57","modified_gmt":"2026-05-13T06:20:57","slug":"perimeter-and-area-ncert-solutions-class-9-ganita-manjari","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/perimeter-and-area-ncert-solutions-class-9-ganita-manjari\/","title":{"rendered":"Perimeter and Area – NCERT Solutions Class 9 Ganita Manjari"},"content":{"rendered":"\n

Measuring Space: Perimeter and Area – NCERT Solutions Class 9 Maths Ganita Manjari includes all the questions with solutions given in NCERT Class 9 Ganita Manjari<\/a> textbook.<\/p>\n\n\n\n

NCERT Solutions Class 9<\/h2>\n\n\nEnglish Kaveri<\/a>\n\n\nHindi Ganga<\/a>\n\n\nSanskrit Sharada<\/a>\n\n\nMaths Ganita Manjari<\/a>\n\n\nScience Exploration<\/a>\n\n\nSocial Understanding Society<\/a>\n\n\n\n

Measuring Space: Perimeter and Area \u2013 NCERT Solutions<\/h2>\n\n\n\n
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Q.1:<\/p>\n\n\n\n

The perimeter of a circle is 44 cm. What is its radius?<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

We use the circumference formula {tex}C=2 \\pi r{\/tex}.
Given {tex}C=44 {~cm}{\/tex} and taking {tex}\\pi \\approx \\frac{22}{7}{\/tex}:<\/p>\n\n\n\n

{tex}44=2 \\times \\frac{22}{7} \\times r{\/tex}
{tex}44=\\frac{44}{7} \\times r{\/tex}
{tex}r=\\frac{44 \\times 7}{44}=7 {~cm}{\/tex}<\/p>\n\n\n\n

The radius of the circle is {tex}{7 \\ c m}{\/tex}.<\/p>\n\n\n\n

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Q.2:<\/p>\n\n\n\n

Calculate, correct to 3 significant figures, the circumference of a circle with: (i) radius 7 cm (ii) radius 10 cm (iii) radius 12 cm.<\/p>\n\n\n\n

Solution:<\/p>\n\n\n\n

    \n
  1. Radius 7 cm: {tex} C=2 \\times \\pi \\times 7 \\approx 43.982 {\/tex} {tex}\\Longrightarrow 44.0 {~cm}{\/tex}<\/li>\n\n\n\n
  2. Radius 10 cm: {tex} C=2 \\times \\pi \\times 10 \\approx 62.831{\/tex} {tex} \\Longrightarrow {6 2 . 8} {~cm}{\/tex}<\/li>\n\n\n\n
  3. Radius 12 cm: {tex} C=2 \\times \\pi \\times 12 \\approx 75.398 {\/tex} {tex}\\Longrightarrow {7 5 . 4} {~cm}{\/tex}<\/li>\n<\/ol>\n\n\n\n
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    Q.3:<\/p>\n\n\n\n

    Calculate the length of the arc of a circle if: (i) the radius is 3.5 cm and the angle at the centre is 60o<\/sup>, and (ii) the radius is 6.3 m and the angle at the centre is 120o<\/sup>.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

      \n
    1. Radius {tex}r=3.5 \\ {cm}{\/tex}, Angle {tex}\\theta=60^{\\circ}{\/tex}
      Step 1: Plug values into the formula: {tex} l=\\frac{60}{360} \\times 2 \\times \\frac{22}{7} \\times 3.5{\/tex} Step 2: Simplify the fraction and the radius {tex}\\left(3.5=\\frac{7}{2}\\right){\/tex}: {tex} l=\\frac{1}{6} \\times 2 \\times \\frac{22}{7} \\times \\frac{7}{2}{\/tex} Step 3: Cancel the common terms (2 and 7): {tex} l=\\frac{1}{6} \\times 22=\\frac{11}{3} \\approx 3.67 \\ {cm}{\/tex}<\/li>\n\n\n\n
    2. Radius {tex}r=6.3 \\ {m}{\/tex}, Angle {tex}\\theta=120^{\\circ}{\/tex}
      Step 1: Plug values into the formula: {tex} l=\\frac{120}{360} \\times 2 \\times \\frac{22}{7} \\times 6.3{\/tex} Step 2: Simplify the fraction {tex}\\left(\\frac{120}{360}=\\frac{1}{3}\\right){\/tex} and the decimal: {tex} l=\\frac{1}{3} \\times 2 \\times \\frac{22}{7} \\times 6.3{\/tex} Step 3: Divide 6.3 by 7 to get 0.9: {tex} l=\\frac{1}{3} \\times 2 \\times 22 \\times 0.9{\/tex} Step 4: Multiply the remaining numbers: {tex} l=\\frac{1}{3} \\times 44 \\times 0.9=44 \\times 0.3={1 3 . 2} {m}{\/tex}<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Q.4:<\/p>\n\n\n\n

      Find the perimeter of a sector (i.e., the curved portion as well as the two straight portions) of a circle of radius 14 cm and sector angle 75o<\/sup>.<\/p>\n\n\n\n

      Solution:<\/p>\n\n\n\n

      Step 1: Calculate the Arc Length (l)
      The formula for the curved part is {tex}l=\\frac{\\theta}{360} \\times 2 \\pi r{\/tex}.
      Using {tex}\\theta=75^{\\circ}, r=14 {~cm}{\/tex}, and {tex}\\pi \\approx \\frac{22}{7}{\/tex}:<\/p>\n\n\n\n

      {tex} l=\\frac{75}{360} \\times 2 \\times \\frac{22}{7} \\times 14{\/tex}<\/p>\n\n\n\n